Nyoj 298 Phase Change point (Matrix high-speed power)

Change time limit for points: -Ms | Memory Limit:65535KB Difficulty:5

Descriptive narrative

There are no more than 10,000 points on a plane. The coordinates are known. Now it's possible to do the following for all points:

Translation a certain distance (M), the relative x-axis upside-down (x), relative to the y-axis flip (y), the coordinates are reduced or enlarged by a certain multiple (S), all points to the coordinates of the origin counter-clockwise rotation of a certain angle (R).

The number of operations is not more than 1 million times, the coordinates of all points are finally obtained.

Tip: Assuming that the value of pi is used in the program, it can be obtained with ACOS (-1.0).

Input

There's just one set of test data
The first line of the test data is two integer n,m, representing the number of points and the number of operations (n<=10000,m<=1000000)
The subsequent row has n logarithm pairs, and the first number of each pair represents the x-coordinate of a point, and the second number represents the y-coordinate. The absolute value of the initial coordinate size of these points is no more than 100.
The subsequent m lines, each representing an operation, are a character at the beginning of the line:
The first character, assuming M, represents a panning operation. The line will be followed by two digits x, Y. Indicates that all points are shifted by vector (x, y);
The first character is assumed to be X. It means that all points are flipped up and down relative to the x-axis;
The first character is assumed to be y, which means that all points are flipped left and right relative to the y-axis;
The first character is assumed to be s. Then it will be followed by a number p, which indicates that the coordinates are enlarged p times;
The first character is assumed to be r, then it is followed by a number a, which means that all points relative to the origin of coordinates counterclockwise rotate a certain angle a (in degrees)

Output

output two numbers per line. Represents the coordinates of a point (rounding the result to 1 digits after the decimal point, and outputting one decimal place)
The output order of the points should be consistent with the input order

Example input

2 51.0 2.0 2.0 3.0XYM 2.0 3.0S 2.0R 180

Example output

-2.0-2.00.0 0.0

Analysis: The hypothesis is modeled according to the description of the topic. Must be timed out. At this point we need to find a high-speed transformation method.

This allows you to calculate the last rectangle formed after the M-transform, and then multiply the coordinates of the point by the rectangle to find out the answer.

#include <cstdio> #include <cstring> #include <cmath> #define PI acos ( -1.0) struct Matrix {double mat[3]    [3];        Matrix () {memset (mat, 0, sizeof (MAT));    for (int i = 0; i < 3; i++) mat[i][i] = 1;        } Matrix Multi (Matrix A, Matrix B) {matrix res;                for (int i = 0, i < 3; i++) {for (int j = 0; J < 3; J + +) {Res.mat[i][j] = 0;                for (int k = 0; k < 3; k++) {res.mat[i][j] = Res.mat[i][j] + a.mat[i][k] * B.mat[k][j];    }}} return res;        } Matrix translation (Matrix A, double p, double q) {//pan up P units, shift Q units to the right matrix res;        RES.MAT[0][2] = p;        RES.MAT[1][2] = q;    Return Multi (res, A);        } Matrix scale (Matrix A, double p) {//zoom P times Matrix Res;        Res.mat[0][0] = res.mat[1][1] = p;    Return Multi (res, A);        } Matrix Turn_ud (Matrix A) {//axes flip The matrix res up or down; res.mat[1][1] =-1;    Return Multi (res, A);        } Matrix TURN_LR (Matrix A) {//axes flip the matrix res around;        Res.mat[0][0] =-1;    Return Multi (res, A);        } Matrix Rotate (Matrix A, double angle) {//rotate counterclockwise around Origin angle angle double rad = angle/180.0 * PI;        Matrix Res; Res.mat[0][0] = cos (RAD);        RES.MAT[0][1] =-sin (RAD); Res.mat[1][0] = sin (rad);        RES.MAT[1][1] = cos (RAD);    Return Multi (res, A); }};struct point {double X, y;}    P[10005];int Main () {int n, m;    Char op[5];    Double x, y;    scanf ("%d%d", &n, &m);    for (int i = 0; i < n; i++) scanf ("%lf%lf", &p[i].x, &AMP;P[I].Y);    Matrix A;        for (int i = 0; i < m; i++) {scanf ("%s", op);        if (op[0] = = ' X ') A = A.turn_ud (a);        else if (op[0] = = ' Y ') A = A.TURN_LR (a);            else if (op[0] = = ' M ') {scanf ("%lf%lf", &x, &y);        A = A.translation (A, x, y); } else if (op[0] = = ' S ') {scanf ("%lf", &x);        A = A.scale (A, x);            } else if (op[0] = = ' R ') {scanf ("%lf", &x);        A = A.rotate (A, x);        }} for (int i = 0; i < n; i++) {double xx = a.mat[0][0] * p[i].x + a.mat[0][1] * p[i].y + a.mat[0][2];        Double yy = a.mat[1][0] * p[i].x + a.mat[1][1] * p[i].y + a.mat[1][2];    printf ("%.1lf%.1lf\n", XX, yy); } return 0;}

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Nyoj 298 Phase Change point (Matrix high-speed power)